So this is in exercise 3.14 (Neat!) in Baby Rudin, which I have found quite a simple and obvious proof for, however when I checked the answers the proofs I found were quite complicated so now I am a bit skeptical of my own proof. So can someone check it out?
The question is:
If $\{s_n\}$ is a complex sequence, define its arithmetic mean $\sigma_n$ by $$ \sigma_n=\frac{s_0+s_1+...+s_n}{n+1}$$ If $\lim s_n=s$ prove that $\lim \sigma_n = s$
Now here is my solution:
For all $\frac{\epsilon}{2}>0$ we can find a positive integer $N$ such that $n\geq N$ implies $|s_n-s|<\frac{\epsilon}{2}$
and thus $|\frac{s_0+...+s_n+...s_n+m}{n+m+1}-s|$=$$|\frac{s_0+...+s_{n-1}-ns_n}{n+m+1}+\frac{(s_n-s)+(s_{n+1}-s)+...+(s_{n+m}-s)}{n+m+1}|$$
$$\leq |\frac{s_0+...+s_{n-1}-ns_n}{n+m+1}|+\frac{(s_n-s)+(s_{n+1}-s)+...+(s_{n+m}-s)}{n+m+1}|$$
Now we can take $m$ to be large enough such that $|\frac{s_0+...+s_{n-1}-ns_n}{n+m+1}|<\frac{\epsilon}{2}$ and thus we have: $$|\frac{s_0+...+s_n+...s_n+m}{n+m+1}-s|<\epsilon$$
and we're done.
Can anyone tell me where the mistake is (if there is any)
